Chapter 3, continued

Chapter 3,
continued
Mixed Review for TEKS (p. 200)
1. B;
Using the data in the table, the equations R 5 35x and
B 5 20x represent the total cost, R, to rent the roller rink
and the total cost, B, to rent the bowling alley, where x is
the number of hours that Marcos rents the facility.
2. H;
The line y 5 23x 1 5 is parallel to the line y 5 23x
because both lines have the same slope, 23. The line
1
y 5 23x 1 5 is perpendicular to the line y 5 }3x 1 5
because the product of their slopes is 21.
4. F;
17
The shortest distance Dana can walk to reach the hiking
path is a path that is perpendicular to the hiking path.
Equation of the hiking path line:
30 2 15
15
3
5}
5 }2
m5}
10 2 0
10
y 5 mx 1 b
3
15 5 }2 (0) 1 b
An equation for the vertical distance is y 5 }
x, where x
25
is the horizontal distance.
17
When x 5 500: y 5 }
(500) 5 340
25
The vertical distance from where you started is 340 feet.
275 2 200
75
5. m 5 } 5 } 5 15
5
520
The slope is 15.
Chapter 3 Review (pp. 202–205)
15 5 b
1. Two lines that do not intersect and are not coplanar are
3
y 5 }2 x 1 15
Slope of the line that is perpendicular to the hiking
path line:
3
2
} + m 5 21
2
m 5 2}3
Slope of the line perpendicular to the hiking path passing
through (21, 14):
y 5 mx 1 b
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3. B;
2
14 5 2}3 (21) 1 b
28 5 b
2
y 5 2}3 x 1 28
The point where the two lines intersect:
2
3
}x 1 15 5 2} x 1 28
3
2
13
} x 5 13
6
x56
called skew lines.
2. Alternate interior angle pairs lie between the two lines
and on opposite sides of the transversal while consecutive
interior angle pairs lie between the two lines and on the
same side of the transversal.
3. Ž1 and Ž5 are corresponding angles.
4. Ž3 and Ž6 are alternate interior angles.
5. Ž4 and Ž6 are consecutive interior angles.
6. Ž7 and Ž2 are alternate interior angles.
7. The equation 14x 2 2y 5 26 is in standard form.
8. The equation y 5 7x 2 13 is in slope-intercept form.
}
9. NR > @##$
QR
}
11. NJ is skew to @##$
QR.
}
10. NP i @##$
QR
12. Plane NJK is parallel to plane LMQ.
13. By the Vertical Angles Congruence Theorem,
mŽ1 5 548. By the Alternate Interior Angles Theorem,
mŽ2 5 mŽ1 5 548.
14. By the Consecutive Interior Angles Theorem,
mŽ1 1 958 5 1808. So, mŽ1 5 858. By the Alternate
Interior Angles Theorem, mŽ2 5 958.
15. By the Corresponding Angles Postulate,
y 5 2}3 (6) 1 28 5 24
mŽ1 5 1358. mŽ1 1 mŽ2 5 1808 because they are
supplementary angles.
The distance between (21, 14) and (6, 24):
1358 1 mŽ2 5 1808
2
}}
d 5 Ï(6 2 21)2 1 (24 2 14)2
mŽ2 5 458
}
5 Ï225 1 100
}
5 Ï325
}
5 5Ï13
The shortest
distance Dana can walk to reach the hiking
}
path is 5Ï 13 feet.
Geometry
Worked-Out Solution Key
79
Chapter 3,
continued
16. By the Corresponding Angles Postulate, y 5 35.
x8 1 358 5 1808 because they are supplementary angles.
x 1 35 5 180
25. a. P(3, 21), y 5 6x 2 4
m56
y 5 mx 1 b
x 5 145
21 5 6(3) 1 b
17. By the Alternate Interior Angles Theorem,
(5x 2 17)8 5 488.
19 5 b
y 5 6x 2 19 is parallel to the given line.
5x 2 17 5 48
1
b. m 5 2}
6
5x 5 65
y 5 mx 1 b
x 5 13
488 1 y8 5 1808 because they are supplementary angles.
48 1 y 5 180
1
21 5 2}6 (3) 1 b
1
2}2 5 b
y 5 132
18. By the Corresponding Angles Postulate, 2y8 5 588.
2y 5 58
1
1
y 5 2}6 x 2 }2 is perpendicular to the given line.
26. a. P(26, 5), 7y 1 4x 5 2
y 5 29
7y 5 24x 1 2
2x8 1 588 5 1808 because they are supplementary
angles.
2x 1 58 5 180
4
4
m 5 2}7
2x 5 122
2
y 5 2}7 x 1 }7
y 5 mx 1 b
x 5 61
4
5 5 2}7(26) 1 b
19. 3 2
1
11
7
}5b
4
11
is parallel to the given line.
y 5 2}7 x 1 }
7
20. x8 1 738 5 1808
21. (x 1 14)8 5 1478
x 5 107
x 5 133
7
23
326
27. The lines have a slope of } 5 } 5 3, so a
21
21 2 0
x 5 32
23. Line 1: (8, 12), (7, 25)
1
217
21
m1 5 } 5 } 5 17
1
m2 5 }
5}
5 2}
17
17
8 2 (29)
The lines are perpendicular.
24. Line 1: (3, 24), (21, 4)
8
m1 5 }
5}
5 22
21 2 3
24
Line 2: (2, 7), (5, 1)
127
6
m2 5 }
5 2}3 5 22
522
The lines are parallel.
80
Geometry
Worked-Out Solution Key
perpendicular line has a slope of 2}3.
The segment from (21, 3) to (2, 2) has a slope of
Line 2: (29, 3), (8, 2)
4 2 (24)
31
y 5 }4 x 1 }
is perpendicular to the given line.
2
5x 5 160
21
7
5 5 }4 (26) 1 b
}5b
5x 1 20 5 180
223
y 5 mx 1 b
31
2
22. (2x 1 20)8 1 3x8 5 1808
25 2 12
728
7
b. m 5 }
4
223
2 2 (21)
1
3
} 5 2}.
So, the distance between the lines is
}}
}
d 5 Ï(2 2 (21))2 1 (2 2 3)2 5 Ï10 ø 3.2 units.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Ž2 and Ž3 are complementary. So, mŽ2 5
908 2 mŽ3 5 908 2 558 5 358. By the Corresponding
Angles Postulate, mŽ1 5 mŽ2 5 358.
Chapter 3,
continued
2
826
28. The lines have a slope of } 5 }, so a
5
3 2 (22)
18. P(3, 5), m 5 28
y 5 mx 1 b
5
perpendicular line has a slope of 2}2.
5 5 28(3) 1 b
The segment from (22, 6) to (0, 1) has a slope of
126
0 2 (22)
y 5 28x 1 29
5
2
} 5 2}.
19. P(1, 3), y 5 2x 2 1
So, the distance between the lines is
}}
1
}
d 5 Ï(0 2 (22))2 1 (1 2 6)2 5 Ï29 ø 5.4 units.
Chapter 3 Test (p. 206)
2m 5 21 l m 5 2}2
y 5 mx 1 b
1
1. Ž1 and Ž8 are alternate exterior angles.
3 5 2}2 (1) 1 b
2. Ž2 and Ž6 are corresponding angles.
}5b
7
2
3. Ž3 and Ž5 are consecutive interior angles.
5. Ž3 and Ž7 are corresponding angles.
20. P(0, 2), y 5 2x 1 3
6. Ž3 and Ž6 are alternate interior angles.
2m 5 21 l m 5 1
7. x 5 140
8. (18x 2 22)8 5 508
9. (4x 1 11)8 5 1078
18x 5 72
4x 5 96
x54
x 5 24
10. x8 1 1378 5 1808
y 5 mx 1 b
2 5 1(0) 1 b
25b
y5x12
21. P(2, 23), x 2 y 5 4
x 5 43
11. x8 5 (128 2 x)8
2y 5 2x 1 4
12. 738 1 (x 1 17)8 5 1808
2x 5 128
x 1 90 5 180
x 5 64
x 5 90
13. (3, 21), (3, 4)
m 5 21
y 5 mx 1 b
21 5 b
y 5 2x 2 1
The slope is undefined.
22. 688 1 x8 5 908
14. (2, 7), (21, 23)
210
y5x24
23 5 21(2) 1 b
4 2 (21)
5
m5}
5 }0
323
23 2 7
10
m5}
5}
5}
21 2 2
23
3
x 5 22
23. 518 1 3x8 5 908
15. (0, 5), (26, 12)
12 2 5
7
1
y 5 2}2 x 1 }2
4. Ž4 and Ž5 are alternate interior angles.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
29 5 b
7
m5}
5 2}6
26 2 0
16. P(22, 4), m 5 3
y 5 mx 1 b
4 5 3(22) 1 b
24. x8 1 (8x 1 9)8 5 908
3x 5 39
9x 5 81
x 5 13
x59
25. (0, 30), (50, 60)
60 2 30
30
3
m5}
5}
5 }5
50 2 0
50
y 5 mx 1 b
3
10 5 b
y 5 3x 1 10
1
17. P(7, 12), m 5 2}
5
y 5 mx 1 b
1
12 5 2}5 (7) 1 b
30 5 }5 (0) 1 b
30 5 b
3
y 5 }5 x 1 30
3
When x 5 100: y 5 }5 (100) 1 30 5 90
The cost of renting the van for a 100-mile trip is $90.
67
}5b
5
1
67
y 5 2}5 x 1 }
5
Geometry
Worked-Out Solution Key
81
Chapter 3,
continued
10. Let x 5 score on fourth quiz.
Chapter 3 Algebra Review (p. 207)
1. y > 22x 1 3
76 1 81 1 77 1 x
4
2. ya0.5x 2 4
}}q80
y
y
1
x
1
234 1 x
4
}q80
234 1 x q 320
1
21
x q 86
x
He must get a score of at least 86 to have an average of
at least 80.
11. 0.07x 1 5 < 0.12x
3. 22.5x 1 yq1.5
yq2.5x 1 1.5
5 < 0.05x
100 < x
y
After 100 minutes of calls, the cost of using Company A
is less than the cost of using Company B.
2
22
TAKS Practice (pp. 210–211)
x
1. C;
f (x) 5 0.8(x 2 2) 1 3.2(x 1 1.5)
y 5 0.8(x 2 2) 1 3.2(x 1 1.5)
5. y < 22
4. x < 3
y 5 0.8x 2 1.6 1 3.2x 1 4.8
y
y
1
21
y 5 4x 1 3.2
x
The y-intercept is 3.2.
1
2. J;
x
1
(0, 1), (2, 22)
22 2 1
7. 2x 1 3yq218
2y > 25x 2 5
3yq22x 2 18
y < 5x 1 5
yq2}3x 2 6
3
The rate of change is 2}2.
2
y
3. C;
y
1
21
x
1
1
3
5 2}2
m5}
220
The y-intercept of y 5 20.25(x 2 8) 5 20.25x 1 2 is
b 5 2. You can see from the graph that the y-intercept
decreases from 2 to 26, while the x-intercept decreases
and the slope remains the same.
y 5 20.25(x 2 8) y
x
4
8. 3x 2 4ya6
(0, 2)
(224, 0)
(8, 0)
24ya23x 1 6
4
y 5 20.25x 2 6
3
yq}3 x 2 }2
x
(0, 26)
4
y
If the y-intercept is changed to 26, the x-intercept
decreases.
1
21
x
4. G;
(1, 21), (0, 5)
5 2 (21)
9. Let x 5 number of months.
8xa46
xa5.75
It will take at most 6 months for Eric to repay his mother.
6
5}
5 26
m5}
021
21
The slope of the line that contains these 3 points is 26.
5. A;
(0, 6), (3, 26)
26 2 6
212
5}
5 24
m5}
320
3
y 5 mx 1 b 5 24x 1 6
82
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
6. 5x 2 y > 25
Chapter 3,
continued
Cumulative Review, Chs. 1–3 (pp. 212–213)
6. H;
If line p is shifted so that the x-intercept decreases and
the y-intercept stays the same, p is not as steep, but it
still rises from left to right. So, the slope decreases and
is positive.
7. D;
Each time the value of x increases by 1, the value of f (x)
decreases by 3. So, f (x) is a linear function whose rate
of change is 23. Because the function is linear, it can be
written in the form f (x) 5 mx 1 b, where m is the rate of
change. Substitute values from the table to find b.
f (x) 5 mx 1 b
2x 5 27
x57
GH 5 5x 2 7 5 5(7) 2 7 5 28
FH 5 2(28) 5 56
2. 3x 2 5 5 x 1 3
2x 5 8
x54
XY 5 3x 2 5 5 3(4) 2 5 5 7
XZ 5 2(7) 5 14
10 5 23(22) 1 b
3. mŽ A 5 288
45b
So, the expression 23x 1 4 can be used to find the
values of f (x) in the table.
4. mŽ A 5 1138
acute
obtuse
5. mŽ A 5 798
6. mŽA 5 908
acute
8. G;
Jay earns $5 per hour, so for h hours of work he earns
5h dollars. This number plus the tip, $10, gives his total
earnings, t. So, the equation is t 5 5h 1 10.
9. B;
right
7. P 5 2* 1 2w 5 2(14) 1 2(6) 5 40
The perimeter is 40 inches.
A 5 *w 5 14(6) 5 84
The area is 84 square inches.
Kate’s age 5 Sue’s age 1 4
8. P 5 a 1 b 1 c 5 15 1 14 1 13 5 42
Kate’s age 1 Sue’s age 5 30
The perimeter is 42 inches.
The pair of equations that represents the situation is
k5s14
k 1 s 5 30
1
1
A 5 }2 bh 5 }2(14)(12) 5 84
The area is 84 square inches.
9. P 5 4s 5 4(3.8) 5 15.2
10. J;
P 5 2(x 1 5) 1 2(2x)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1. 4x 5 5x 2 7
The perimeter is 15.2 yards.
5 2x 1 10 1 4x
A 5 s 2 5 (3.8)2 5 14.44
5 6x 1 10
The area is 14.44 square yards.
11. B;
10. The numbers are all perfect cubes. The next number
22x 1 y 5 1
is 125.
y 5 2x 1 1
11. Each of the numbers is divided by 4 to create the next
1
number. The next number is }2 .
A line parallel to 22x 1 y 5 1 has a slope of 2.
P(2, 6), m 5 2
12. Each number is multiplied by 23 to create the next
y 5 mx 1 b
number. The next number is 162.
6 5 2(2) 1 b
13. Because 6x 5 24 and 24 < 42, you know x < 7.
25b
14. Because mŽ A 5 1038 and 1038 > 908, you know ŽA is
y 5 2x 1 2
an obtuse angle.
12. G; The eye colors of the people who responded “Other”
16. 3x 2 14 5 34
13. B;
6x5y22
6
10
3
5
2
3x
5y z
}
5 }x5 2 3y22 2 4z21 5 }x2y26z21 5 }
6
3 4
10x y z
15. Because the musician is playing a violin, the musician is
playing a stringed instrument.
cannot be determined from the graph.
14. The events are independent.
P(red and then even) 5 P(red) + P(even)
5 4
5 1
5
ø 0.313
5 }8 + }8 5 }8 + }2 5 }
16
Given
3x 5 48
Addition Property of Equality
x 5 16
Division Property of Equality
17. 24(x 1 3) 5 228
x1357
x54
Given
Division Property of Equality
Subtraction Property of Equality
The probability that the spinner lands on red on the
first spin and on an even number on the second spin is
about 0.313.
Geometry
Worked-Out Solution Key
83
continued
18. 43 2 9(x 2 7) 5 2x 2 6
1
b. m 5 2}
6
Given
43 2 9x 1 63 5 2x 2 6
Distributive Property
29x 1 106 5 2x 2 6
Simplify.
28x 1 106 5 26
Addition Property of
Equality
28x 5 2112
1
Subtraction Property of
Equality
x 5 14
y 5 mx 1 b
Division Property of
Equality
19. 348 1 (4x 1 30)8 5 1808
22 5 2}6 (3) 1 b
3
2}2 5 b
3
1
y 5 2}6 x 2 }2
29. P(22, 12), y 5 2x 2 3
a. m 5 21
y 5 mx 1 b
4x 1 64 5 180
12 5 21(22) 1 b
4x 5 116
10 5 b
x 5 29
y 5 2x 1 10
20. 4x8 1 (7x 1 37)8 5 1808
b. m 5 1
11x 1 37 5 180
y 5 mx 1 b
11x 5 143
12 5 1(22) 1 b
x 5 13
21. (9x 1 54)8 1 5x8 5 1808
14 5 b
14x 5 126
y 5 x 1 14
x59
30. P(7, 21), 6y 1 2x 5 18
(3y 1 42)8 5 (9x 1 54)8
6y 5 22x 1 18
3y 1 42 5 9(9) 1 54
1
y 5 2}3x 1 3
3y 1 42 5 135
3y 5 93
1
a. m 5 2}
3
y 5 31
y 5 mx 1 b
22. (5x 2 10)8 5 (4x 1 4)8
1
21 5 2}3 (7) 1 b
x 5 14
3y8 5 (5y 2 80)8
4
3
}5b
22y 5 280
1
y 5 40
23.
y 5 79
x8 1 y8 5 1808
x 1 79 5 180
x 5 101
25. (5, 22), (7, 22)
22 2 (22)
m5}
50
725
27. (21, 2), (0, 4)
422
m5}
52
0 2 (21)
28. P(3, 22), y 5 6x 1 7
a. m 5 6
y 5 mx 1 b
22 5 6(3) 1 b
220 5 b
y 5 6x 2 20
24.
7y8 5 1408
y 5 20
x8 1 1408 5 1808
x 5 40
b. m 5 3
y 5 mx 1 b
21 5 3(7) 1 b
222 5 b
26. (8, 3), (3, 14)
14 2 3
4
y 5 2}3 x 1 }3
11
m5}
5 2}
5
328
y 5 3x 2 22
31. Yes, @##$
AC > @##$
DB, because if two lines intersect to form
a linear pair of congruent angles, then the lines are
perpendicular.
32. The photograph suggests a line perpendicular to a plane.
33. The photograph suggests parallel and
perpendicular lines.
34. The photograph suggests parallel and
perpendicular planes.
35. The distance from Westville to Reading is
37 1 52 5 89 miles.
36. A 5 *w 5 40(25) 5 1000
The area of the garden is 1000 square feet.
84
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 3,
Chapter 3,
continued
37. If you want the lowest prices on new televisions, then
come and see Matt’s TV Warehouse.
Hypothesis: You want the lowest prices on
new televisions.
Conclusion: Come and see Matt’s TV Warehouse.
38. Converse: If you come and see Matt’s TV Warehouse,
then you want the lowest prices on new televisions.
Inverse: If you do not want the lowest prices on
new televisions, then do not come and see Matt’s TV
Warehouse.
Contrapositive: If you do not come and see Matt’s TV
Warehouse, then you do not want the lowest prices on
new televisions.
39. Yes, the last plank is the same length as the first because
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
of the Transitive Property of Congruence of Segments.
Geometry
Worked-Out Solution Key
85